\(xy^2-\left(x-2\right)\left(x^4+2x+1\right)=2y^2\)
\(\Rightarrow xy^2-2y^2-\left(x-2\right)\left(x^4+2x+1\right)=0\)
\(\Rightarrow y^2\left(x-2\right)-\left(x-2\right)\left(x^4+2x+1\right)=0\)
\(\Rightarrow\left(x-2\right)\left(y^2-x^4-2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\y^2-x^4-2x-1=0\end{matrix}\right.\)
Thay \(x=2\) vào \(y^2-x^4-2x-1=0\) ta có:
\(y^2-2^4-2\cdot2-1=0\)
\(\Rightarrow y^2-21=0\)
\(\Rightarrow y^2=21\)
\(\Rightarrow\left[{}\begin{matrix}y=\sqrt{21}\\y=-\sqrt{21}\end{matrix}\right.\)
Vậy (x;y) thỏa mãn là: \(\left(2;\sqrt{21}\right);\left(2;-\sqrt{21}\right)\)
lý thuyết đầy đủ các phuong phap giai phuong trinh nghiem nguyen
